Problem 15 - Chemistry The Molecular Science

Question

Write equilibrium constant expressions for these reactions: (a) \(\mathrm{CH}_{4}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{~g})\) (b) \(4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (c) \(\mathrm{BaCO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{BaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})\) (d) \(\mathrm{NH}_{3}(\mathrm{~g})+\mathrm{HCl}(\mathrm{g}) \rightleftharpoons \mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s})\)

Step-by-Step Solution

Answer

(a) \( K_c = \frac{[\mathrm{CO}][\mathrm{H}_2]^3}{[\mathrm{CH}_4]} \); (b) \( K_c = \frac{[\mathrm{NO}]^4[\mathrm{H}_2\mathrm{O}]^6}{[\mathrm{NH}_3]^4[\mathrm{O}_2]^5} \); (c) \( K_c = [\mathrm{CO}_2] \); (d) \( K_c = \frac{1}{[\mathrm{NH}_3][\mathrm{HCl}]} \).

Step 1: Identifying the phases

For equilibrium constant expressions, only gases (g) and aqueous solutions (aq) are included. Solids (s) and liquids (l) are excluded.

Step 2: Writing expression for reaction (a)

The reaction is: \( \mathrm{CH}_{4}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{~g}) \). Since \( \mathrm{H}_2 \mathrm{O} \) is a liquid, it is not included. The equilibrium expression is: \[ K_c = \frac{[\mathrm{CO}][\mathrm{H}_2]^3}{[\mathrm{CH}_4]} \]

Step 3: Writing expression for reaction (b)

The reaction is: \( 4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \). All substances are gases. The equilibrium expression is: \[ K_c = \frac{[\mathrm{NO}]^4[\mathrm{H}_2\mathrm{O}]^6}{[\mathrm{NH}_3]^4[\mathrm{O}_2]^5} \]

Step 4: Writing expression for reaction (c)

The reaction is: \( \mathrm{BaCO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{BaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g}) \). Solids \( \mathrm{BaCO}_{3} \) and \( \mathrm{BaO} \) are not included. The equilibrium expression is: \[ K_c = [\mathrm{CO}_2] \]

Step 5: Writing expression for reaction (d)

The reaction is: \( \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{HCl}(\mathrm{g}) \rightleftharpoons \mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s}) \). The solid \( \mathrm{NH}_4 \mathrm{Cl} \) is not included. The equilibrium expression is: \[ K_c = \frac{1}{[\mathrm{NH}_3][\mathrm{HCl}]} \]

Key Concepts

Equilibrium Constant Expressions

The equilibrium constant expression, denoted as \( K_c \), is a fundamental concept in understanding chemical equilibrium. It provides a quantitative measure of the concentration of products and reactants at equilibrium. For any given reaction at equilibrium, the concentrations of the gases or aqueous solutions are used to express this balance. Solid and liquid phases are excluded as their concentrations remain constant and don't affect the equilibrium condition.

• For a generic reaction: \( aA + bB \rightleftharpoons cC + dD \), the equilibrium constant expression is given by:

\[ K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \]The coefficients in the balanced chemical equation become the exponents in the expression, reflecting the stoichiometry of the reaction. This calculation helps chemists predict the direction of the reaction and its extent, providing insight into the balance between reactants and products.

Phases of Matter in Equilibrium

Matter can exist in different phases: solid, liquid, and gas. In the context of equilibrium reactions, only the gaseous and aqueous phases are considered in the equilibrium constant expression. This is because these states can change concentration during chemical reactions, while solids and liquids do not.

Gases and solutions change in concentration and affect the reaction's balance, which is why they are included in \( K_c \).

Solids and liquids are not included because their densities/mass per unit volume are constant during the reaction.

This inclusion and exclusion are fundamental when writing equilibrium constant expressions, ensuring that only the components that actively participate in changing equilibrium are considered.

Exclusion of Solids and Liquids

In equilibrium constant expressions, we exclude pure solids and pure liquids. The reasoning behind this stems from their constancy in concentration or density. This means: - Solids: Their concentration doesn't change because they have a fixed volume and mass. - Liquids: Especially pure liquids, like water in many reactions, have a constant concentration and do not affect the equilibrium. Therefore, the reaction's equilibrium is represented only by the substances whose concentrations can change, thus simplifying the expression without losing accuracy. This principle allows for clear calculations and predictions about reaction behavior and shifts in equilibrium.

Gas Phase Reactions

Reactions where gases act as reactants or products are particularly interesting in equilibrium studies because the concentrations can change dramatically with pressure and volume. In gas-phase reactions, all participating substances are gaseous, and the equilibrium constant expression includes all reactants and products because their concentrations change.For example, in the reaction of \( 4 \mathrm{NH}_3 \) with \( 5 \mathrm{O}_2 \): \[ K_c = \frac{[\mathrm{NO}]^4[\mathrm{H}_2\mathrm{O}]^6}{[\mathrm{NH}_3]^4[\mathrm{O}_2]^5} \]Here, the calculation reflects the balanced equation coefficients. By understanding this, we can predict the outcome of reactions under various conditions, such as pressure changes affecting \( K_p \) (a version of \( K_c \) used when pressure is involved), making such knowledge applicable in industries such as chemical manufacturing and environmental science.